Saturday, July 1, 2017

fx - Pricing and Hedging an Option through a Currency Triangle


How is the option price of an plain vanilla option (in a Black Scholes setting) derived, which is written on, say XAGGBP but practically hedged with XAGUSD and GBPUSD (because these are more liquid)? Eventually, I am interested in the delta(s) and correlation risk with respect to XAGUSD and GBPUSD.



Answer



Let $X_t^{xag\rightarrow gbp}$ be the exchange rate from one unit of XAG to units of GBP. Moreover, let $X_t^{xag\rightarrow usd}$, and $X_t^{gbp\rightarrow usd}$ be the respective exchanges rates from one unit of XAG and GBP to units of USD. We consider an option payoff, in GBP, at maturity $T$ of the form \begin{align*} \left(X_T^{xag\rightarrow gbp} -K\right)^+. \tag{1} \end{align*} Note that \begin{align*} \left(X_T^{xag\rightarrow gbp} -K\right)^+ = \left(\frac{X_T^{xag\rightarrow usd}}{X_T^{gbp\rightarrow usd}} -K\right)^+. \end{align*} We assume that, under the USD risk-neutral probability measure $Q_{usd}$, \begin{align*} dX_t^{xag\rightarrow usd} &= X_t^{xag\rightarrow usd}\left[\left(r^{usd}-r^{xag} \right)dt +\sigma_1 dW_t^1 \right],\\ dX_t^{gbp\rightarrow usd} &= X_t^{gbp\rightarrow usd}\left[\left(r^{usd}-r^{gbp} \right)dt +\sigma_2\left(\rho dW_t^1 +\sqrt{1-\rho^2}dW_t^2\right)\right], \end{align*} where $r^{usd}$, $r^{gbp}$, and $r^{xag}$ are interest rates, $\sigma_1$ and $\sigma_2$ are volatilities, $\rho$ is the correlation, and $\{W_t^1, \, t\ge 0\}$ and $\{W_t^2, \, t\ge 0\}$ are two standard independent Brownian motions.


Let $B_t^{usd}=e^{r^{usd} t}$ and $B_t^{gbp}=e^{r^{gbp} t}$ be the respective USD and GBP money market account values at time $t$. Moreover, let $Q^{gbp}$ be the GBP risk-neutral probability measure. Note that \begin{align*} \frac{dQ^{gbp}}{dQ^{usd}}\big|_t &= \frac{B_t^{gbp}X_t^{gbp\rightarrow USD}}{B_t^{usd}X_0^{gbp\rightarrow USD}}\\ &=e^{-\frac{1}{2}\sigma_2^2 t + \sigma_2\left(\rho W_t^1 +\sqrt{1-\rho^2}W_t^2\right)}. \end{align*} Then, $\{\tilde{W}_t^1, \, t\ge 0\}$ and $\{\tilde{W}_t^2, \, t\ge 0\}$, where \begin{align*} \tilde{W}_t^1 &= W_t^1 - \sigma_2\rho t, \\ \tilde{W}_t^2 &= W_t^2 - \sigma_2\sqrt{1-\rho^2} t, \end{align*} are two standard independent Brownian motions under $Q_{gbp}$. Furthermore, under $Q^{gbp}$, \begin{align*} dX_t^{xag\rightarrow usd} &= X_t^{xag\rightarrow usd}\left[\left(r^{usd}-r^{xag} +\rho\sigma_1\sigma_2\right)dt +\sigma_1 d\tilde{W}_t^1 \right],\\ dX_t^{gbp\rightarrow usd} &= X_t^{gbp\rightarrow usd}\left[\left(r^{usd}-r^{gbp} +\sigma_2^2\right)dt +\sigma_2\left(\rho d\tilde{W}_t^1 +\sqrt{1-\rho^2}d\tilde{W}_t^2\right)\right]. \end{align*} Then \begin{align*} X_t^{xag\rightarrow gbp} &= \frac{X_t^{xag\rightarrow usd}}{X_t^{gbp\rightarrow usd}} \\ &=\frac{X_0^{xag\rightarrow usd}}{X_0^{gbp\rightarrow usd}} e^{\left(r^{gbp}-r^{xag}+\rho\sigma_1\sigma_2-\frac{1}{2}\sigma_1^2 -\frac{1}{2}\sigma_2^2\right)t + (\sigma_1-\rho\sigma_2)\tilde{W}_t^1 -\sigma_2\sqrt{1-\rho^2}\tilde{W}_t^2}\\ &=X_0^{xag\rightarrow gbp} e^{\left(r^{gbp}-r^{xag}-\frac{\sigma_1^2+\sigma_2^2-2\rho\sigma_1\sigma_2}{2}\right)t + \sqrt{\sigma_1^2+\sigma_2^2 - 2\rho\sigma_1\sigma_2 }\frac{(\sigma_1-\rho\sigma_2)\tilde{W}_t^1 -\sigma_2\sqrt{1-\rho^2}\tilde{W}_t^2}{\sqrt{\sigma_1^2+\sigma_2^2 - 2\rho\sigma_1\sigma_2 }} }. \end{align*}


Let \begin{align*} \sigma = \sqrt{\sigma_1^2+\sigma_2^2 - 2\rho\sigma_1\sigma_2 }, \end{align*} and \begin{align*} W_t^3 = \frac{(\sigma_1-\rho\sigma_2)\tilde{W}_t^1 -\sigma_2\sqrt{1-\rho^2}\tilde{W}_t^2}{\sqrt{\sigma_1^2+\sigma_2^2 - 2\rho\sigma_1\sigma_2 }}. \end{align*} Then $\{W_t^3, \, t \ge 0\}$ is a standard Brownian motion under $Q^{gbp}$, by Levy's characterization. Moreover, \begin{align*} d X_t^{xag\rightarrow gbp} = X_t^{xag\rightarrow gbp}\left[\left(r^{gbp}-r^{xag} \right)dt +\sigma dW_t^3 \right]. \end{align*} Therefore, the option payoff $(1)$ can be valued using the Garman Kohlhagen formula, while replace the initial exchange rate $X_0^{xag\rightarrow gbp}$ by $\frac{X_0^{xag\rightarrow usd}}{X_0^{gbp\rightarrow usd}}$. The respective hedge ratios can be computed subsequently.


No comments:

Post a Comment

technique - How credible is wikipedia?

I understand that this question relates more to wikipedia than it does writing but... If I was going to use wikipedia for a source for a res...